18 minutes ago, dalestephenson said:

If you place the first 2 damage on Thalin, you reduce the number of hero killing shadows from an undefended attack to 3 instead of 5. It's still higher than 5% (3/26 rounds to 12%). In order to get below 5% you need to place the first 2 damage on Thalin and *defend* with Gimli, which reduces the number of killing Shadows to 1 (1/26).

No, the point is that you can put either undefended attack on Thalin if there isn't a bad shadow. It doesn't have to be the first. You need both shadows to be "bad" (or one to be Hummerhorns) to lose a hero with my strategy. This makes the risk less than 5% when you consider the fact that in half the cases where one of the shadows is Hummerhorns, you would have defended the wrong attack with Gimli anyway. So a first-order calculation of the risk is around:

%hummerhorns/2 + %bad*%bad = 1/27/2 + 4/27 * 4/27 = 4%

Edit: actually since there are two shadows the chance of seeing Hummerhorns is 2/27, not 1/27, so the risk is closer to 5.9%.

Edit 2: But when you include the consideration that there are combinations that are loses no matter what you do, like Hummerhorns/Orcs and Orcs/Orcs, this drops even further.

Seastan, what you seem to be missing is that without Citadel Plate, Gimli has only 2 hp remaining -- he *cannot* take an undefended attack without the Plate. If you take the first undefended attack and it discards the plate, you *have* to put it on Thalin to avoid hero death, and then have a second attack to take that will kill a hero if you take it undefended. So it goes like this:

1st attack (taken undefended)

1/27 -- Eowyn dies instantly (Hummerhorns)

2/27 -- Gimli takes 5 damage (Dol Goldur Orcs)

2/27 -- Gimli loses plate, Thalin takes 2 damage

22/27 -- no damage shadow [put on Thalin for highest survivability]

2nd attack after no damage shadow (taken undefended)

1/26 -- Eowyn and Thalin die instantly (Hummerhorns)

2/26 -- Gimli takes 5 damage (Dol Goldur Orcs)

2/26 -- Gimli loses plate, choose a hero to take undefended and die

21/26 -- Gimli takes two damage.

2nd attack after Dol Goldur shadow (taken undefended)

1/26 -- Eowyn and Gimli die instantly (Hummerhorns)

1/26 -- 5 damage attack, choose a hero to die (Dol Goldur Orcs)

2/26 -- Gimli loses plate and dies, Thalin can take damage

22/26 -- Thalin takes two damage

2nd attack after Plate losing shadow (taken undefended)

Choose a hero to die, all die with Hummerhorns

2nd attack after Plate losing shadow (Gimli defends)

1/26 -- all three heroes die (Hummerhorns)

2/26 -- Gimli dies (Dol Goldur Orcs)

23/36 -- No hero dies, but Gimli can't counterattack and you still have two engaged enemies

So if you think that the second attack *must* be taken undefended, since removing an enemy is essential even if it costs Thalin's life, the odds off hero death work out to be:

1/27 [hummerhorns] + 2/27 [DGO] * 4/26 + 2/27 [lost plate] + 22/27 * 3/26 == 21.65%

However, if you want to defend with Gimli after losing the plate during the first attack, you can reduce the odds to this:

1/27 [hummerhorns] + 2/27 [DGO] * 4/26 + 2/27[lost plate] * 3/26 + 22/27 * 3/26 = 15.1%

I think it makes more sense to sacrifice Thalin, but even if you don't you're still at 15.1% overall. If you have no bad shadow on the first undefended attack, place the damage on Thalin, and elect to take the second attack undefended the odds of hero death from *that* attack are 11.5%, and that's after something *didn't* go wrong on the first attack.

I grant that I wasn't considering the loss of the plate, and was treating the non-Hummerhorns shadows as equally bad, when they're not.

But you're still missing a major factor in this discussion. The relevant question is not "What are the odds of a hero dying if I take these two attacks undefended", but "What are increased odds of a hero dying if I take these two attacks undefended". For that we need to know the probability that you'd lose a hero even if Gimli had defended.

Let's say the first attack is undefended, and Gimli defends the second attack.

1st attack (taken undefended)

1/27 -- Eowyn dies instantly (Hummerhorns)

2/27 -- Gimli takes 5 damage (Dol Goldur Orcs)

1/27 -- Gimli loses plate, Thalin takes 2 damage (Cavern)

1/27 -- Gimli loses plate, Thalin takes 2 damage (Driven)

22/27 -- no damage shadow [put on Thalin for highest survivability]

2nd attack after no damage shadow (Gimli defends)

1/26 -- 1 damage each, Thalin dies (Hummerhorns)

2/26 -- Gimli takes 1 damage (Dol Goldur Orcs)

1/26 -- Gimli loses plate

1/26 -- Eowyn loses key

21/26 -- no damage shadow

2nd attack after Dol Goldur shadow (Gimli defends)

1/26 -- 1 damage each, Thalin dies (Hummerhorns)

1/26 -- +1 attack, Gimli dies (Dol Goldur Orcs)

1/26 -- Gimli loses plate, dies

1/26 -- Eowyn loses key

22/26 -- no damage shadow

2nd attack after Cavern (Gimli defends)

1/26 -- 1 damage each, kills Thalin (Hummerhorns)

2/26 -- +1 attack (Dol Goldur Orcs)

1/26 -- Gimli loses plate

22/26 -- no damage shadow

2nd attack after Driven (Gimli defends)

1/26 -- 1 damage each, kills Thalin (Hummerhorns)

2/26 -- +1 attack (Dol Goldur Orcs)

23/26 -- no damage shadow

----

Total probability of hero death: 1/27 [Hum] + (22/27 [no damage] * 1/26 [Hum]) + (2/27 [DGO] * 3/26) + (1/27 [Cavern] * 1/26 [Hum]) + (1/27 [Driven] * 1/26 [Hum]) = 8.0%

So there was a 8.0% chance you were going to die no matter what, and that certainly matters in assessing the risk of a certain strategy.

If you are really committed to taking both undefended, no matter what the first shadow effect is, then the increased risk of losing a hero by doing so would be 21.65% (from your calculation)-8.0% = 13.65%

On the other hand, you point out that if the first shadow removes Gimli's plate, taking the second undefended is a guaranteed hero loss, so I'd probably defend. In this case the increased risk of the strategy is 15.1% (from your calculation)-8.0% = 7.1%, not far off my original 5% estimate.

Edit: missed a condition

4 hours ago, sappidus said:

I suspect analysis of the remaining cases would be best done using Markov chains, the mere appearance of which would probably narrow the interested public to statistically 0%, heh.

Ha, first I need to figure out, what a Markov chain is.

47 minutes ago, Amicus Draconis said:

Ha, first I need to figure out, what a Markov chain is.

I looked at Wikipedia, felt dumb, closed it and put 3 copies of hasty stroke in my deck. Just in case...

2 hours ago, Seastan said:

So there was a 8.0% chance you were going to die no matter what, and that certainly matters in assessing the risk of a certain strategy.

If you are really committed to taking both undefended, no matter what the first shadow effect is, then the increased risk of losing a hero by doing so would be 21.65% (from your calculation)-8.0% = 13.65%

On the other hand, you point out that if the first shadow removes Gimli's plate, taking the second undefended is a guaranteed hero loss, so I'd probably defend. In this case the increased risk of the strategy is 15.1% (from your calculation)-8.0% = 7.1%, not far off my original 5% estimate.

With decreased risk comes decreased results -- if you defend with Gimli, you end the turn with two engaged enemies. However, more to the point is that if you defend with Gimli when you remove plate, you *aren't* actually taking a second undefended attack at all, and the question was specifically about the risk from taking two undefended attacks.

Still that's a bit afar afield from the original discussion about TaGimli, given that no one would *ever* take an undefended attack for the purpose of powering up Gimli if it were guaranteed to kill a hero! The reasoning behind killing Thalin would be to let an already-powered-up-Gimli one-shot an enemy, which is an entirely different strategic decision. The true decision point is after the shadow on the first undefended was revealed as harmless (OK, other than massively raising their threat). Then the possibilities are:

1) Put the two damage on Gimli (+2 attack), and defending with Gimli -- 3/26 chance of hero death, 2/26 chance of Gimli adding +3 attack.

2) Put the two damage on Thalin and defending with Gimli -- 1/26 chance of hero death, 2/26 chance of Gimli adding +1 attack.

3) Put the two damage on Thalin and taking the attack undefended -- 3/26 chance of hero death, 2/26 chance of Gimli +5 attack, 21/26 chance of Gimli +2 attack, and Gimli ready to kill one of the enemies.

4) Put the two damage on Gimli (+2 attack) and taking the attack undefended -- 5/26 chance of hero death.

Beorn's actual decision was 15% worse than the safest option, but it's far better than the safest option because the safest option doesn't kill an enemy and doesn't ramp up Gimli's damage. It's not better than #3 strategically just because the extra damage Beorn got from putting everything on Gimli *wasn't needed* to eliminate an enemy this turn.

Suppose the shadow situation was the same, and the attackers were attacking for 4 and 2. If you take the 2 undefended first, that effectively eliminates #1 and #4 as options. It raises the death rate to 3/26 defending and 5/26 undefended -- the same relative margin, but gives a benefit the current #2 does not have -- it is guaranteed to increase Gimli's attack by at least two. But it has the same problem -- Gimli can't strike back and eliminate an enemy. In this case *despite* the additional risk, it is sensible to take an undefended 4-strength attack. (You could take them in the opposite order and take the 4 undefended first, but it's the same situation -- any bad shadow on the 4-strength kills, and if you defend the 2-strength you can't kill an enemy). That's purely a theoretical consideration, though. The only 4-attack enemy is the Nazgul, and his magic one-shot number is 12, which even Gimli needs a lot of help to get to. (He did get to 8 by the end of the round, the minimum necessary to one shot the already-engaged Dungeon Jailor and the minimum to two-shot the Nazgul.)

22 minutes ago, dalestephenson said:

It's not better than #3 strategically just because the extra damage Beorn got from putting everything on Gimli *wasn't needed* to eliminate an enemy this turn.

I agree. But tying to figure out the effect on the game win percentage is way too had, which is why I was careful to limit my statement to increased risk of hero loss, not game loss.

I think it's safe to say that we are far enough away from the topic of Thorin here that we should move further discussion of Gimli (unless it pertains to Thorin) to a new topic.

6 hours ago, Amicus Draconis said:

This post alone took me about 3 hours to compute and write down

Hence why I went for the simulation route 😁

5 hours ago, Halberto said:

I looked at Wikipedia, felt dumb, closed it and put 3 copies of hasty stroke in my deck. Just in case...

About the same for me. Need to look at it, when I am really awake this afternoon.

2 hours ago, Amicus Draconis said:

About the same for me. Need to look at it, when I am really awake this afternoon.

Markov chains are a way of simplifying problems by ignoring history and psychology; they are a favoured technique of mathematicians and engineers.

14 hours ago, Seastan said:

Hence why I went for the simulation route 😁

As I have not programmed for years, that would probably take even longer for me. But I needed most of the three hours to remember how combinations work, double check for mistakes and then write it down in a hopefully readable manner. I guess for some of the other cases I will not need as much time. By the way, what sequence did you use with Thorin? Discarding first and then drawing / using Master of the Forge or the other way round? I would assume it makes no difference, but I better ask anyway.

7 hours ago, ColinEdwards said:

Markov chains are a way of simplifying problems by ignoring history and psychology; they are a favoured technique of mathematicians and engineers.

I hope they will simplify this problem for me as well .

4 minutes ago, Amicus Draconis said:

As I have not programmed for years, that would probably take even longer for me. But I needed most of the three hours to remember how combinations work, double check for mistakes and then write it down in a hopefully readable manner. I guess for some of the other cases I will not need as much time. By the way, what sequence did you use with Thorin? Discarding first and then drawing / using Master of the Forge or the other way round? I would assume it makes no difference, but I better ask anyway.

I hope they will simplify this problem for me as well .

I always programmed Thorin last, as he'd probably be discarding in the combat phase after drawing or using Master of the Forge. Not that it makes any difference to the result.

24 minutes ago, Seastan said:

I always programmed Thorin last, as he'd probably be discarding in the combat phase after drawing or using Master of the Forge. Not that it makes any difference to the result.

Another question: Are there any other attachments in your deck, which the Master of the Forge can draw out? I assume not.

Edit: Or at least you do not have to draw any besides Steward.

40 minutes ago, Amicus Draconis said:

Another question: Are there any other attachments in your deck, which the Master of the Forge can draw out? I assume not.

Edit: Or at least you do not have to draw any besides Steward.

Actually I assumed if you're running Master of the Forge you have more attachments than just Steward. So I was generous and assumed that Master of the Forge always found some attachment.

Thanks, that should help.

My first results were:

On 11/20/2018 at 6:47 PM, Amicus Draconis said:

p1 = 3 / 44 = 6.8%.

p2 = 41 / 44 * 3 / 43 = 6.5%

p3 = (1 - p1 - p2) * 3 / 42 = 6.2%

p4 = (1 - p1 - p2 - p3) * 3 / 41 = 5.9%

p5 = (1 - p1 - p2 - p3 - p4) * 3 / 40 = 5.6%

p6 = (1 - p1 - p2 - p3 - p4 - p5) * 3 / 39 = 5.3%

p7 = (1 - p1 - p2 - p3 - p4 - p5 - p6) * 3 / 38 = 5.0%

E = (6.8% * 1 + 6.5% * 2 + 6.2% * 3 + 5.9% * 4 + 5.6% * 5 + 5.3% * 6 + 5.0% * 7) / 41.3% = 3.8

This time, I will have a look two other cases:

2a) Extra card each round, with no Thorin

This now means, over the course of 7 rounds, a total of 14 cards will be drawn. As one can see, the individual probabilities form a mathematical sequence, which allows to compute the next seven steps. I could write the sequence down, but this forum is not suited to do so and thus it looks terrible:

pn = (1 - sum (pm, 1..n..m-1) * 3 / (45 - n)

p8 = 4.8%

p9 = 4.5%

p10 = 4.2%

p11 = 4.0%

p12 = 3.7%

p13 = 3.5%

p14 = 3.3%

The sum of all 14 probabilities yields 69.3% which is also the result of 1 - C(3, 0) *C(41, 14) / C(44, 14) and slightly lower than Seastan's result of 69.5%.

The expected value this time looks as follows, slightly higher than Seastan's result of 3.5:

E = ((6.8% + 6.5%) * 1 + (6.2% + 5.9%) * 2 + (5.6% + 5.3%) * 3 + (5.0% +4.8%) * 4 + (4.5% + 4.2%) * 5 + (4.0% + 3.7%) * 6 + (3.5% + 3.3%) * 7) / 69.3% = 3.6

Seastan's simulation was a little bit more generous with drawing Steward, but as mentioned before, there always is a small variance in such simulations. Off to the next case:

1b) Drawing 1 card per turn, with Thorin each round

Let's define a function p(n, m) = n / m which gives the probability of Steward being the top card of a deck containing m cards, when n Stewards are left in the deck. Every even card will be drawn and every odd one will be discarded by Thorin.

The probability for having Steward on top of the 44 card deck with 3 Stewards in it and thus drawing it in the first round is the same as p1: p(3, 44) = 3 / 44 = 6.8%. Of course there is a 1 - p(3, 44) = 93.2% chance, this will not be the case. Hereafter we will only look at cases, that have not yet been won, as there is no need to draw or discard any more cards, when the first Steward is already in hand.

In the first round Thorin will discard a Steward with the probability of p(3, 43) = 3 / 43 = 7.0% which will only happen in (1 - p(3, 44)) * p(3, 43) = 6.5% of all games. Again this is the same probability as p2, after all the first Steward is the second card of the deck in this case. But now it becomes more interesting, as discarding Stewards from the deck not only makes the deck smaller, it also reduces the chance for another Steward. Not discarding a Steward at this point will happen with a (1 - p(3, 44)) * (1 - p(3, 43)) = 86.7% chance in all games.

The draw in the second round depends on whether one of our Stewards has already been discarded or not. If not, then there is a p(3, 42) = 7.1% chance of a Steward on top of the deck. Multiplying both relevant probabilities yields (1 - p(3, 44)) * (1 - p(3, 43)) * p(3, 42) = 6.2%, which equals p3. But we are only in the second round and we have yet to calculate, how often we will draw a Steward in the second round, when one copy has been discarded in the first.

If a Steward has already been discarded, the chance of another one on top of the deck at this point is p(2, 42) = 4.8%. This will only happen in (1 - p(3, 44)) * p(3, 43) * p(2, 42) = 0.3% of all games. Adding both probabilities of drawing a Steward in the second round yields 6.2% + 0.3% = 6.5% which equals p2. So no matter what card has been discarded in the first round, the probability to draw a Steward in the second round is the same, as when not discarding a card at all.

This nice exercise can now be repeated for all 7 rounds (I will not do it here, as it would be far too long for this post), and the end result will show, that discarding will not affect your chances of seeing a copy of Steward (as long as you do not run out of your deck). If anyone is interested in my OpenOffice spreadsheet, feel free to ask.

And as the probabilities are the same as in case 1a), the expected value will also be the same.

Thorin Stonehelm does not say (once per round) so what is preventing a player from just discarding multiple cards to get multiple bonuses?

You can only use a Response once per qualifying trigger, so you can only use Thorin's response once per attack. You can use it again for another attack in the same round if you ready Thorin and attack a different enemy.

If you could use a Response as many times as you want for a single trigger, you could, for example, trigger Horn of Gondor infinitely and get infinite resources any time 1 ally is destroyed.